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Clock Hands Trisecting Face?

Date: 04/05/99 at 13:28:39
From: Ken Logan
Subject: Problem Solving (Time problems) Graduate Level

Dear Dr. Math,

I would appreciate your help on the following problem. I have asked 
the same question once already but have not heard a response from you.

Question: At what time (if any) do the three hands of a clock trisect 
its face?

Here is my solution: There are 60 1-minute spaces around a clock face.  
The rate of the second hand is 60 spaces per minute, the rate of the 
minute hand is 1 space per minute, and the rate of the hour hand is 
1/12 space per minute.

We begin by letting all three hands begin at the 12:00 position. We 
let Sh represent the number of spaces the hour hand moves until the 
hands trisect the face, Sm the number of spaces the minute hands 
move, and Ss the number of spaces the second hand moves. By the 
distance = rate x time formula we have Ss = 60T, Sm = T and 
Sh = (1/12)T. Notice that the time it takes each hand to be equal 
spaces apart is the same. Furthermore, Ss = Sm + 1200 and Ss = Sh + 
14,400, and because Sm = T we have Ss = T + 1200. By substitution we 
have T + 1200 = 60T. 

Solving this equation for T we have T = 20 20/59 minutes. This is the 
number of minutes until the second and minute hands are 20 minutes 
apart. Where do I go from here? Do I take an entirely different 

Thank you,
Ken Logan

Date: 04/05/99 at 16:53:44
From: Doctor Anthony
Subject: Re: Problem Solving (Time problems) Graduate Level

It looks as if the three hands never exactly trisect the clock face.

In 12 hours    Hour hand moves 360 degrees
               Minute hand moves 12 x 360 degrees
               Second hand moves 12 x 60 x 360 = 720 x 360 degrees

and these are in the ratio  1 : 12 : 720

If a = the angle moved through by hour hand, then 12a = the angle 
moved through by the minute hand and 720a = the angle moved through by 
the second hand.

So we require    12a - a = 120
                     11a = 120  so a = 10.90909.... degrees
   and               12a = 130.90909.......

    But  720a = 7854.5454.....

subtracting whole revolutions from this we get 7854.5454 - 7560
                                              = 294.5454...

But  294.5454 - 130.90909 = 163.6363 which does NOT equal 120 degrees.

If we instead have  12a - a = 240
                        11a = 240
                          a = 21.81818

and then  12a = 261.81818

and 720a = 15709.09091   and subtracting whole revolutions 

           15709.09091 - 15480 = 229.0909

and this is clearly not 120 degrees from 261.81818

So in both cases we cannot get the hands to exactly trisect the face.

- Doctor Anthony, The Math Forum   
Associated Topics:
High School Puzzles

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